Question

Normals are drawn at point $P,Q,R$ lying on parabola $y^2=4x$ which intersect at $(3,0)$ then area of $△PQR$ is :-

• A. $4$
• B. $2$
• C. $1$
• D. $Noneofthese$

Answer 1

The correct option is B $2$

$y^2=4x$

$2y$ $dy/dx=4$

Slope of tangent $=2/y$

Slope of normal$=-y/2$

Equation of normal

$y=(-y/2)(x-3)$

$2=3-x$

$x=1$ or $y=0$

Points P, Q, R are $(0,0),(1,2)$ & $(1,-2)$

Base $QR=4$

Height $=1$

Area $=0.5\ast 4\ast 1$

$=2$.